Question

Sketch the graph of the given parametric equations; using a graphing utility is advisable. Be sure to indicate the orientation of the graph. \(x=2 \sec t, \quad y=3 \tan t, \quad-\pi / 2

Short answer

The graph is a hyperbola with orientation from bottom right to top right.

Step-by-step solution

Understand the Parametric Equations

The given parametric equations are \(x = 2 \sec t\) and \(y = 3 \tan t\), with \(-\frac{\pi}{2} < t < \frac{\pi}{2}\). This means for each value of \(t\), we can find a corresponding \(x\) and \(y\). The range of \(t\) is between \(-\frac{\pi}{2}\) and \(\frac{\pi}{2}\), where both \(\sec t\) and \(\tan t\) are defined and exhibit vertical asymptotes.

Identify the Relationship Between x and y

To sketch the graph, let's find a direct relationship between \(x\) and \(y\). We know \(\sec t = \frac{1}{\cos t}\) and \(\tan t = \frac{\sin t}{\cos t}\). Therefore, \(x = 2 \cdot \frac{1}{\cos t}\) and \(y = 3 \cdot \frac{\sin t}{\cos t}\). From \(x\), \(\cos t = \frac{2}{x}\). Substitute in the equation for \(y\): \(y = 3 \cdot \frac{\sin t}{\cos t}\) becomes \(y = 3 \cdot \tan t = 3 \cdot \frac{\sqrt{x^2 - 4}}{x}\) by using \( \sin^2 t = 1 - \cos^2 t = 1 - \left(\frac{2}{x}\right)^2\).

Sketch the Graph

Using a graphing utility, plot the equation \(y = 3 \cdot \frac{\sqrt{x^2 - 4}}{x}\). The graph will be a hyperbola, as it is derived from the hyperbolic identity for secant and tangent. As \(t\) increases from \(-\frac{\pi}{2}\) to \(\frac{\pi}{2}\), both \(x\) and \(y\) traverse the hyperbola, starting from the lower right quadrant to the upper right quadrant of the coordinate plane.

Indicate Orientation

The orientation of the graph is from the bottom right to the top right of the hyperbola. As \(t\) progresses from \(-\frac{\pi}{2}\) to \(\frac{\pi}{2}\), \(x\) approaches the vertical asymptotes at \(x = 2\) and \(x = -2\) (due to \(\sec t\) having asymptotes where \(\cos t = 0\)), while \(y\) traverses the hyperbola, indicating direction.

Key concepts

Secant and Tangent Functions

When working with parametric equations that involve trigonometric functions like secant and tangent, understanding their properties is crucial. The secant function, represented as \( \sec t = \frac{1}{\cos t} \), describes how many times a particular angle wraps around the unit circle in the horizontal direction. It can take any real number value, except where \( \cos t = 0 \), leading to vertical asymptotes in its graph. On the other hand, the tangent function, \( \tan t = \frac{\sin t}{\cos t} \), describes the ratio of the vertical to horizontal components of a circle at a given angle, showing the slope of the line that touches the curve. Understanding these functions helps in sketching curves generated by parametric equations that involve angles, such as those given in the exercise: \( x = 2 \sec t \) and \( y = 3 \tan t \).

Asymptotes

Asymptotes are lines that approach a curve but never actually touch or intersect it. In the context of our parametric equations, \( x = 2 \sec t \) and \( y = 3 \tan t \), vertical asymptotes occur at intervals where \( \sec t \) and \( \tan t \) are undefined. Specifically, this happens when \( \cos t = 0 \) in the formulas for secant and tangent, which are at points \( t = -\frac{\pi}{2} \) and \( t = \frac{\pi}{2} \). These vertical asymptotes show up in the coordinate plane as lines \( x = 2 \) and \( x = -2 \). Recognizing these asymptotes is key to understanding the boundaries and behavior of the hyperbolic curve, as the function's values become very large and infinite near these lines.

Hyperbola Orientation

The orientation of a hyperbola refers to the direction in which the curve is drawn as parameter \( t \) changes. For the parametric equations \( x = 2 \sec t \) and \( y = 3 \tan t \), the hyperbola graph covers the right side of the coordinate plane. As \( t \) moves from \(-\frac{\pi}{2}\) to \(\frac{\pi}{2}\), we observe that the hyperbola travels from the lower right quadrant to the upper right quadrant. This direction is crucial for correctly interpreting the path of the curve. On a graph, the path from lower to upper quadrants indicates the specific trajectory influenced by the parameter \( t \). This trajectory is characteristic of hyperbolas formed by these types of trigonometric parametric equations, emphasizing the role of each value of \( t \) in shaping the curve.

Trigonometric Identities

Trigonometric identities are equations involving trigonometric functions that are universally true for all values of the involved variables. In this exercise, simplifying the relationship between \( x \) and \( y \) was key for graphing. We use the identities \( \sec t = \frac{1}{\cos t} \) and \( \tan t = \frac{\sin t}{\cos t} \) to transform the parametric equations into a singular form: \( y = 3 \cdot \frac{\sqrt{x^2 - 4}}{x} \). This transformation relies on the Pythagorean identity \( \sin^2 t + \cos^2 t = 1 \), which helps turn trigonometric expressions into algebraic forms. Understanding and applying these identities is crucial for converting complex trigonometric parametric expressions into easier-to-handle equations, thus simplifying the sketching of their graphs. By mastering these identities, students can efficiently handle transformations in parametric equations and explore their implications on the graph's shape and trajectory.